Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $24$ units long $\overline{AB}$ is $26$ units long What is $\cot(\angle BAC)?$ $A$ $C$ $B$ $10$ $24$ $26$
Solution: $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)}$ How can we find $\tan(\angle BAC)$ SOH CAH TOA angent = pposite over djacent Opposite $= \overline{BC} = 24$ Adjacent $= \overline{AC} = 10$ $\tan(\angle BAC) = \dfrac{24}{10}$ $\cot(\angle BAC) = \dfrac{1}{\tan(\angle BAC)} = \dfrac{10}{24}$